What is Topology of a metric space?

Example 4.4 (Interior, closure, boundary on one set)

In $\R$, let $A = \intoc{0}{1} \cup \{2\}$. Interior: $\intoo{0}{1}$ — around any $x \in \intoo01$ a small ball stays in $A$; around $1$, every ball $\intoo{1-r}{1+r}$ leaks out of $A$ on the right, so $1$ is not interior; and the isolated $2$ is not interior either. Closure: $\intcc{0}{1} \cup \{2\}$ (the point $0$ is a limit of $A$, nothing else is added). Boundary (closure minus interior): $\{0, 1, 2\}$. Note the asymmetries worth remembering: an endpoint can belong to a set without being interior ($1$), can be adherent without belonging ($0$), and an isolated point is its own boundary ($2$). The same bookkeeping runs verbatim in any metric space, with balls in place of intervals.

Example 4.10 (Open and closed sets recognized by continuity)

The global characterization (Theorem 4.6) is the everyday tool for topological bookkeeping. In $\R^2$: the set $\{(x, y) : x^2 + y^2 < 1,\ y > x^3\}$ is open — it is $g^{-1}(\intoo{-\infty}{1}) \cap h^{-1}(\intoo{0}{+\infty})$ for the continuous $g(x,y) = x^2 + y^2$ and $h(x, y) = y - x^3$, an intersection of two open preimages. In $\bigl(C(\intcc01), d_\infty\bigr)$: the set of functions with $f(0) = f(1)$ and $\int_0^1 f = 0$ is closed — the preimage of $\{(0,0)\}$ under the continuous map $f \mapsto \bigl(f(0) - f(1),\ \int_0^1 f\bigr)$ into $\R^2$ (each coordinate is $1$-Lipschitz, as in Exercise 4.3). The method never draws a picture: exhibit a continuous map, read the set as a preimage, quote the theorem.

Example 4.22 (Reading compactness on covers)

The half-open interval $\intoc{0}{1}$ is covered by the open sets $U_n = \intoo{\frac1n}{2}$, $n \geq 1$; any finite subfamily has a largest index $N$ and misses $\intoc{0}{\frac1N}$: no finite subcover, so $\intoc{0}{1}$ is not compact — which the sequential definition sees through $x_n = \frac1n$, whose limit $0$ escapes. On the other hand, adding the single point $0$ repairs both diagnoses at once: on $\intcc{0}{1}$ every such cover must contain a set containing $0$, which swallows a whole initial segment, and finitely many sets finish the rest. The two languages of Theorem 4.20 always fail or succeed together — covers detect escape exactly where sequences do.